leetcode Evaluate Division

EvaluateDivision

题目详情：

Equations are given in the format

A / B = k

, where

A

and

B

are variables represented as strings, and

k

is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return

-1.0

.

Example:

Given

a / b = 2.0, b / c = 3.0.

queries are:

a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .

return

[6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is:

vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries

, where

equations.size() == values.size()

, and the values are positive. This represents the equations. Return

vector<double>

.

According to the example above:

equations = [ ["a", "b"], ["b", "c"] ],
values = [2.0, 3.0],
queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].

The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

解题方法：

看到这个题目时，不知道应该用什么数据结构来存储图。该开始想的使用map<pair<string, string>, double>，但是这样是不行的，没办法存储每个顶点对应的边。所以参考了discuss中的方法，使用unordered_map<string, unordered_map<string, double> > 来存储图。键值为顶点，每个顶点对应一个map这个map中存储了顶点对应的分母，以及顶点/分母得到的值，这个图是双向的，注意这里要判断，分子是否为0，顶点/分母是否为0。对于每个querie，graph[
4000
a].find(b)
!= graph[a].end() 即能找到以a为分子，b为分母的边，则将结果加入vector。如果找不到以a为分子，b为分母的边，则便利每个a的邻居，看是否能找到以a的邻居为分子，b为分母的边。这里要记录是否便利过某个顶点，没办法使用一个bool的数组来存储这个信息，这个维护一个set，如果便利过，则将其加入set。如过不再set中则没便利过。

代码详情：

class Solution {

public:

    vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {

        unordered_map<string, unordered_map<string, double> > graph;

        vector<double> result;

        for (int i = 0; i < values.size(); i++) {

            graph[equations[i].first].insert(make_pair(equations[i].second, values[i]));

            if (values[i] != 0) {

                graph[equations[i].second].insert(make_pair(equations[i].first, 1 / values[i]));

            }

        }

        for (int i = 0; i < queries.size(); i++) {

            unordered_set<string> s; // to know whether a node is find or not

            double temp = calculate(queries[i].first, queries[i].second, graph, s);

            result.push_back(temp);

        }

        return result;

    }

    double calculate(string a, string b, unordered_map<string, unordered_map<string, double> >& graph, unordered_set<string>& s) {

        if (graph[a].find(b) != graph[a].end()) return graph[a][b];

        unordered_map<string, double>::iterator iter;

        for (iter = graph[a].begin(); iter != graph[a].end(); iter++) {  // calculate neibour of a to find b

            if (s.find((*iter).first) == s.end()) {

                s.insert((*iter).first);

                double temp = calculate((*iter).first, b, graph, s);

                if (temp != -1.0) return (*iter).second * temp;

            }

        }

        return -1.0;

    }

};